[tex]Area: \ A=350 \ m^2 \\ \\ permeter: \ P = 90 \ m \\length: \ l \\ width : \ \ w \\ area : \ A=lw \\ perimeter : \ P=2l+2w \\\\\begin{cases}350=l\cdot w \\ 90=2l+2w \ \ / | \ divide \ each \ term \ by \ 2 \end{cases}[/tex]
[tex]\begin{cases}350=l\cdot w \\ 45= l+ w \end{cases}\\ \\\begin{cases}350=l\cdot w \\ l=45-w \end{cases}\\ \\ substitution \\\\350=(45-w) \cdot w\\ 350= 45w-w^2[/tex]
[tex]w^2-45w+350=0 \\\\w_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{45-\sqrt{(-45)^2-4 \cdot 1\cdot 350}}{2 }=\frac{45-\sqrt{2025-1400}}{2 }=\\\\= \frac{45-\sqrt{625}}{2 }= \frac{45-25}{2 }=\frac{20}{2}=10\\\\w_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{45+\sqrt{(-45)^2-4 \cdot 1\cdot 350}}{2 }= \frac{45-\sqrt{625}}{2 }= \frac{45+25}{2 }=\frac{70}{2}=35[/tex]
[tex]\begin{cases}w=10 \\l=45-w \end{cases} \ \ \ or \ \ \ \begin{cases}w=35 \\l=45-w \end{cases}\\\\\begin{cases}w=10 \\l=45-10 \end{cases} \ \ \ or \ \ \ \begin{cases}w=35 \\l=45-35 \end{cases}\\\\\begin{cases}w=10 \\l=35 \end{cases} \ \ \ or \ \ \ \begin{cases}w=35 \\l=10 \end{cases}\\\\Answer: \ \ \ Since \ this \ is \ a \ pathway, \ it \ makes \ sense \ that \ the \ width \ is \ the \ smaller \\ dimension, \ giving \ a \ width \ of \ 10 \ m \ and \ a \ length \ of \ 35\ m[/tex]
